We denote by $\mathscr { R }$ the set of totally real numbers and we admit that there exists a function $t : \mathscr { R } \rightarrow \mathbb { Q }$ satisfying the following two properties: (i) for $x , y \in \mathscr { R }$ and $\lambda , \mu \in \mathbb { Q }$, we have $t ( \lambda x + \mu y ) = \lambda t ( x ) + \mu t ( y )$ (ii) for $x$ totally positive, we have $t ( x ) \geqslant 0$ and the equality is strict if $x \neq 0$.
We consider a non-zero totally real number $z$. We write $Z ( X ) = X ^ { d } - \left( a _ { d - 1 } X ^ { d - 1 } + \cdots + a _ { 1 } X + a _ { 0 } \right)$ with $d$ minimal. We consider the matrix $S$ of size $d \times d$ whose coefficient $( i , j )$ equals $t \left( z ^ { i + j } \right)$. For $X , Y \in \mathbb { R } ^ { d }$, we set $B ( X , Y ) = X ^ { T } S Y$.
Show that $B$ is an inner product on $\mathbb { R } ^ { d }$.

We denote by $\mathscr { R }$ the set of totally real numbers and we admit that there exists a function $t : \mathscr { R } \rightarrow \mathbb { Q }$ satisfying the following two properties: (i) for $x , y \in \mathscr { R }$ and $\lambda , \mu \in \mathbb { Q }$, we have $t ( \lambda x + \mu y ) = \lambda t ( x ) + \mu t ( y )$ (ii) for $x$ totally positive, we have $t ( x ) \geqslant 0$ and the equality is strict if $x \neq 0$.
We consider a non-zero totally real number $z$. We write $Z ( X ) = X ^ { d } - \left( a _ { d - 1 } X ^ { d - 1 } + \cdots + a _ { 1 } X + a _ { 0 } \right)$ with $d$ minimal. We consider the matrix $S$ of size $d \times d$ whose coefficient $( i , j )$ equals $t \left( z ^ { i + j } \right)$. For $X , Y \in \mathbb { R } ^ { d }$, we set $B ( X , Y ) = X ^ { T } S Y$.
15a. Show that there exists a basis $\left( e _ { 1 } , \ldots , e _ { d } \right)$ of $\mathbb { R } ^ { d }$ with $e _ { i } \in \mathbb { Q } ^ { d }$ for all $i$ and $B \left( e _ { i } , e _ { j } \right) = 0$ for $i \neq j$.
15b. Deduce that there exist $P \in \mathrm { GL } _ { d } ( \mathbb { Q } )$ and $q _ { 1 } , \ldots , q _ { d } \in \mathbb { Q } , q _ { i } > 0$, such that: $$S = P ^ { T } \cdot \operatorname { Diag } \left( q _ { 1 } , \ldots , q _ { d } \right) \cdot P$$

Show that for all matrices $A$ and $B$ in $\operatorname{Sym}^+(p)$ and all non-negative real numbers $a$ and $b$, we have $aA + bB \in \operatorname{Sym}^+(p)$.

Show that if $v \in \mathbf{R}^p$ then the matrix $A = \left(A_{ij}\right)_{(i,j) \in \llbracket 1,p \rrbracket^2}$ defined by $A = vv^T$ is in $\mathrm{Sym}^+(p)$.

Let $d \in \mathbf{N}_*$. We consider a $p$-tuple $\left(x_i\right)_{1 \leq i \leq p}$ of elements of $\mathbf{R}^d$ and the matrix $$A = \left(\left\langle x_i, x_j \right\rangle\right)_{(i,j) \in \llbracket 1,p \rrbracket^2}$$ where $\langle a, b \rangle$ denotes the usual inner product between two vectors $a$ and $b$ of $\mathbf{R}^d$. We denote $|a| = \sqrt{\langle a, a \rangle}$ the norm of $a$.
(a) Show that $A \in \operatorname{Sym}^+(p)$.
(b) We denote $u \in \mathbf{R}^p$ the vector with coordinates $\left(\exp\left(-\frac{|x_1|^2}{2}\right), \ldots, \exp\left(-\frac{|x_p|^2}{2}\right)\right)$. Show that $\left(\exp[A] \odot \left(uu^T\right)\right)_{ij} = \exp\left(-\frac{|x_i - x_j|^2}{2}\right)$ for all $(i,j) \in \llbracket 1,p \rrbracket^2$.
(c) Let $\lambda > 0$ and $K \in \mathcal{M}_p(\mathbf{R})$ the matrix defined by $K_{ij} = \exp\left(-\frac{|x_i - x_j|^2}{2\lambda}\right)$ for all $(i,j) \in \llbracket 1,p \rrbracket^2$. Show that $K \in \operatorname{Sym}^+(p)$.

We fix two $p$-tuples $(x_i)_{i \in \llbracket 1,p \rrbracket}$ and $(a_i)_{i \in \llbracket 1,p \rrbracket}$ of real numbers with the $x_i$ pairwise distinct. We use the notation $\mathcal{S}$, $J$, $J_*$, $\mathcal{S}_*$, $(\mid)_{\mathcal{H}}$ as defined previously.
Let $\mathcal{H}_0 = \{h \in \mathcal{H} \mid h(x_i) = 0\ \forall i \in \llbracket 1,p \rrbracket\}$ and $\tilde{h} \in \mathcal{S}_*$ (we assume here $\mathcal{S}_*$ non-empty).
Show that $\left(\tilde{h} \mid h_0\right)_{\mathcal{H}} = 0$ for all $h_0 \in \mathcal{H}_0$.

We fix two $p$-tuples $(x_i)_{i \in \llbracket 1,p \rrbracket}$ and $(a_i)_{i \in \llbracket 1,p \rrbracket}$ of real numbers with the $x_i$ pairwise distinct. We use the notation $\mathcal{S}$, $\mathcal{S}_*$, $\mathcal{H}_0$, $(\mid)_{\mathcal{H}}$, $\gamma_{2\lambda}$, $\tau_x$ as defined previously. We denote $\mathcal{H}_0^\perp = \{h \in \mathcal{H} \mid \forall h_0 \in \mathcal{H}_0\ (h \mid h_0)_{\mathcal{H}} = 0\}$ the orthogonal subspace to $\mathcal{H}_0$ in $\mathcal{H}$.
(a) Show that $\mathcal{S}_* = \mathcal{S} \cap \mathcal{H}_0^\perp$.
(b) Show that $\mathcal{H}_0^\perp$ contains the vector subspace of $\mathcal{H}$ spanned by the functions $\tau_{x_i}(\gamma_{2\lambda})$ for $i \in \llbracket 1,p \rrbracket$.

We consider $n$ discrete random variables $Y_1, \ldots, Y_n$ with random vector $Y$ and covariance matrix $\Sigma_Y$. The objective is to show that $\mathbb{P}\left(Y - \mathbb{E}(Y) \in \operatorname{Im}\Sigma_Y\right) = 1$. We now assume $r < n$ where $r$ is the rank of $\Sigma_Y$.
Prove that the kernel and image of $\Sigma_Y$ are supplementary orthogonal subspaces in $\mathcal{M}_{n,1}(\mathbb{R})$.

We assume that $\Sigma_Y$ satisfies $$\forall i \in \llbracket 1, n \rrbracket, \quad \sigma_{i,i} = \sigma^2 \quad \text{and} \quad \forall (i,j) \in \llbracket 1, n \rrbracket^2, \quad i \neq j \Longrightarrow \sigma_{i,j} = \sigma^2 \gamma$$ where $\sigma$ and $\gamma$ are two strictly positive real numbers. We denote by $J \in \mathcal{M}_n(\mathbb{R})$ the matrix whose coefficients are all equal to 1.
Specify a unit vector $U_0$ such that the variance of $Z = U_0^\top Y$ is maximal.

Let $V$ and $V^{\prime}$ be two subspaces of $E$ of dimension $p$ (we recall that $p \geqslant 1$).
(a) Show that there exist $u_1 \in V$ and $u_1^{\prime} \in V^{\prime}$ of norm 1 such that $$\left\langle u_1, u_1^{\prime}\right\rangle = \sup\left\{\left\langle a, a^{\prime}\right\rangle \mid\left(a, a^{\prime}\right) \in V \times V^{\prime},\|a\|=\left\|a^{\prime}\right\|=1\right\}.$$
(b) Extend this result by showing that there exist a family $u = (u_1, \ldots, u_p)$ of vectors of $V$ and a family $u^{\prime} = (u_1^{\prime}, \ldots, u_p^{\prime})$ of vectors of $V^{\prime}$ such that $u$ and $u^{\prime}$ are orthonormal and satisfy the following two conditions:
(i) For $k = 1$, we have $$\left\langle u_1, u_1^{\prime}\right\rangle = \sup\left\{\left\langle a, a^{\prime}\right\rangle \mid\left(a, a^{\prime}\right) \in V \times V^{\prime},\|a\|=\left\|a^{\prime}\right\|=1\right\}.$$
(ii) For $k \in \llbracket 2, p \rrbracket$, we have $$\left\langle u_k, u_k^{\prime}\right\rangle = \sup\left\{\left\langle a, a^{\prime}\right\rangle \mid\left(a, a^{\prime}\right) \in V \times V^{\prime},\|a\|=\left\|a^{\prime}\right\|=1,\right.$$ $$\left.\left\langle a, u_l\right\rangle=\left\langle a^{\prime}, u_l^{\prime}\right\rangle=0 \text{ for all } l \in \llbracket 1, k-1 \rrbracket\right\}.$$
(Hint: One may construct the vectors $u_k$ and $u_k^{\prime}$ by induction on the integer $k$.)

Let $V$ and $V^{\prime}$ be two subspaces of $E$ of dimension $p$. Let $u = (u_1, \ldots, u_p)$ and $u^{\prime} = (u_1^{\prime}, \ldots, u_p^{\prime})$ be the orthonormal families constructed in question (1).
(a) Show that $u$ is an orthonormal basis of $V$.
(b) Show that for $k \in \llbracket 1, p-1 \rrbracket$, we have $u_k^{\prime} \in \operatorname{Vect}(u_{k+1}, \ldots, u_p)^{\perp}$. (Hint: One may consider the map $t \mapsto u_k(t) = \frac{u_k + t u_l}{\|u_k + t u_l\|}$ for all $t \in \mathbb{R}$ and $l \in \llbracket k+1, p \rrbracket$ as well as its derivative.)
(c) Show that $u_{k+1} \in \left(\operatorname{Vect}(u_1, \ldots, u_k) + \operatorname{Vect}(u_1^{\prime}, \ldots, u_k^{\prime})\right)^{\perp}$ for all $k$ element of $\llbracket 1, p-1 \rrbracket$.
(d) Deduce that the subspaces $W_k = \operatorname{Vect}(u_k, u_k^{\prime})$ for $k \in \llbracket 1, p \rrbracket$ are pairwise orthogonal.

Let $e = (e_1, \ldots, e_d)$ be an orthonormal basis of $E$, $p$ an integer such that $1 \leqslant p \leqslant d$ and $\mathcal{I}_p = \{\alpha = (i_1, \ldots, i_p) \in \mathbb{N}^p \mid 1 \leqslant i_1 < \cdots < i_p \leqslant d\}$. For all $\alpha = (i_1, \ldots, i_p) \in \mathcal{I}_p$, we denote $e_{\alpha} = (e_{i_1}, \ldots, e_{i_p}) \in E^p$ and for all $\omega$ and $\omega^{\prime}$ elements of $\mathscr{A}_p(E, \mathbb{R})$ $$\langle\omega, \omega^{\prime}\rangle = \sum_{\alpha \in \mathcal{I}_p} \omega(e_{\alpha}) \omega^{\prime}(e_{\alpha}).$$
Show that the inner product $(\omega, \omega^{\prime}) \mapsto \langle\omega, \omega^{\prime}\rangle$ defined above depends only on the inner product on $E$ and not on the choice of the orthonormal basis $e$.

Let $V$ and $V^{\prime}$ be two subspaces of $E$ of dimension $p$ (we recall that $p \geqslant 1$).
(a) Show that there exist $u_1 \in V$ and $u_1^{\prime} \in V^{\prime}$ of norm 1 such that $$\left\langle u_1, u_1^{\prime}\right\rangle = \sup\left\{\left\langle a, a^{\prime}\right\rangle \mid\left(a, a^{\prime}\right) \in V \times V^{\prime},\|a\|=\left\|a^{\prime}\right\|=1\right\}.$$
(b) Extend this result by showing that there exist a family $u = (u_1, \ldots, u_p)$ of vectors of $V$ and a family $u^{\prime} = (u_1^{\prime}, \ldots, u_p^{\prime})$ of vectors of $V^{\prime}$ such that $u$ and $u^{\prime}$ are orthonormal and satisfy the following two conditions:
(i) For $k = 1$, we have $$\left\langle u_1, u_1^{\prime}\right\rangle = \sup\left\{\left\langle a, a^{\prime}\right\rangle \mid\left(a, a^{\prime}\right) \in V \times V^{\prime},\|a\|=\left\|a^{\prime}\right\|=1\right\}.$$
(ii) For $k \in \llbracket 2, p \rrbracket$, we have $$\left\langle u_k, u_k^{\prime}\right\rangle = \sup\left\{\left\langle a, a^{\prime}\right\rangle \mid\left(a, a^{\prime}\right) \in V \times V^{\prime},\|a\|=\left\|a^{\prime}\right\|=1,\right.$$ $$\left.\left\langle a, u_l\right\rangle=\left\langle a^{\prime}, u_l^{\prime}\right\rangle=0 \text{ for all } l \in \llbracket 1, k-1 \rrbracket\right\}.$$
(Hint: One may construct the vectors $u_k$ and $u_k^{\prime}$ by induction on the integer $k$.)

We fix two orthonormal families $u = (u_1, \ldots, u_p)$ of vectors of $V$ and $u^{\prime} = (u_1^{\prime}, \ldots, u_p^{\prime})$ of vectors of $V^{\prime}$ satisfying the conditions of question (1).
Show that if $\operatorname{dim}(V \cap V^{\prime}) \geqslant 1$, then $u_k = u_k^{\prime}$ for all $1 \leqslant k \leqslant \operatorname{dim}(V \cap V^{\prime})$.

We fix two orthonormal families $u = (u_1, \ldots, u_p)$ of vectors of $V$ and $u^{\prime} = (u_1^{\prime}, \ldots, u_p^{\prime})$ of vectors of $V^{\prime}$ satisfying the conditions of question (1).
(a) Show that $u$ is an orthonormal basis of $V$.
(b) Show that for $k \in \llbracket 1, p-1 \rrbracket$, we have $u_k^{\prime} \in \operatorname{Vect}(u_{k+1}, \ldots, u_p)^{\perp}$. (Hint: one may consider the map $t \mapsto u_k(t) = \frac{u_k + u_{\ell}}{\|u_k + t u_{\ell}\|}$ for all $t \in \mathbb{R}$ and $\ell \in \llbracket k+1, p \rrbracket$ as well as its derivative.)
(c) Show that $u_{k+1} \in \left(\operatorname{Vect}(u_1, \ldots, u_k) + \operatorname{Vect}(u_1^{\prime}, \ldots, u_k^{\prime})\right)^{\perp}$ for all $k$ element of $\llbracket 1, p-1 \rrbracket$.
(d) Deduce that the subspaces $W_k = \operatorname{Vect}(u_k, u_k^{\prime})$ for $k \in \llbracket 1, p \rrbracket$ are pairwise orthogonal.

Let $e = (e_1, \ldots, e_d)$ be an orthonormal basis of $E$, $p$ an integer such that $1 \leqslant p \leqslant d$ and $\mathcal{I}_p = \{\alpha = (i_1, \ldots, i_p) \in \mathbb{N}^p \mid 1 \leqslant i_1 < \cdots < i_p \leqslant d\}$. For all $\omega$ and $\omega^{\prime}$ elements of $\mathcal{A}_p(E, \mathbb{R})$: $$\langle\omega, \omega^{\prime}\rangle = \sum_{\alpha \in \mathcal{I}_p} \omega(e_{\alpha}) \omega^{\prime}(e_{\alpha}) \tag{1}$$
Show that the inner product $(\omega, \omega^{\prime}) \mapsto \langle\omega, \omega^{\prime}\rangle$ defined by (1) depends only on the inner product on $E$ and not on the choice of the orthonormal basis $e$.

Deduce that if $v\in\mathcal{H}$, then the restriction of $B$ to $v^\perp$ is an inner product.

Let $A \in \mathcal{M}_{m,n}(\mathbb{R})$ be a matrix with $m$ rows and $n$ columns. We denote by $\langle u, v \rangle_{\mathbb{R}^n}$ the inner product between two vectors $u$ and $v$ of $\mathbb{R}^n$ and $\langle \mu, \nu \rangle_{\mathbb{R}^m}$ that between two vectors $\mu$ and $\nu$ of $\mathbb{R}^m$.
(a) Show that for all $(x, \nu) \in \mathbb{R}^n \times \mathbb{R}^m$, we have $$\langle Ax, \nu \rangle_{\mathbb{R}^m} = \left\langle x, A^\top \nu \right\rangle_{\mathbb{R}^n},$$ where $A^\top$ denotes the transpose matrix of $A$.
(b) Deduce that $\ker A \subset (\operatorname{Im} A^\top)^\perp$ where $E^\perp$ denotes the orthogonal complement of $E$ for the inner product on $\mathbb{R}^n$ for any vector subspace $E$ of $\mathbb{R}^n$.
(c) Show that $\ker A = (\operatorname{Im} A^\top)^\perp$.

We denote by $\mathbf{e}$ the matrix of $\mathcal{M}_{n,1}(\mathbb{R})$ whose coefficients are all equal to 1, and $P$ the matrix of order $n$ defined by $$P = I_n - \frac{1}{n} \mathbf{e} \cdot \mathbf{e}^T.$$
Show that $P$ is symmetric and that the endomorphism of $\mathbf{R}^n$ canonically associated is an orthogonal projection onto $\operatorname{Vect}(\mathbf{e})^\perp$.

Let $u = (u_k)_{k \geqslant 0}$ be a sequence of $\mathbb{C}$ such that $\mathbb{M}_n(u) \neq \emptyset$. Let $A \in \mathbb{M}_n(u)$. We assume throughout this part that $A$ is diagonalizable in $\mathscr{M}_n(\mathbb{C})$ and we denote by $\lambda_1, \cdots, \lambda_\ell$ its eigenvalues with $\lambda_i \neq \lambda_j$ if $i \neq j$. For every $k \in \llbracket 1; \ell \rrbracket$ we define the polynomial: $$Q_k^A(X) = \prod_{j=1, j\neq k}^{\ell} \frac{X - \lambda_j}{\lambda_k - \lambda_j}$$
(a) Show that $$u(A) = \sum_{k=1}^{\ell} U(\lambda_k) Q_k^A(A).$$
(b) Show that for every $k \in \llbracket 1; \ell \rrbracket$, $Q_k^A(A)$ is a projection whose image and kernel we will specify.
(c) Deduce that $$\sum_{k=1}^{\ell} Q_k^A(A) = I_n.$$

Let $Z \in \mathscr{M}_{d}(\mathbb{R})$ be an invertible matrix with singular value decomposition $Z = VDU^{T}$ where $U = (u_{1}|\ldots|u_{d})$, $V = (v_{1}|\ldots|v_{d})$ are orthogonal matrices and $D = \operatorname{Diag}(\sqrt{\lambda_{1}}, \ldots, \sqrt{\lambda_{d}})$. We assume that $\operatorname{det}(Z) > 0$.

• [(a)] Show that if $R \in \mathrm{SO}_{d}(\mathbb{R})$ then $V^{T}RU \in \mathrm{SO}_{d}(\mathbb{R})$.
• [(b)] Show that $$\sup_{R \in \mathrm{SO}_{d}(\mathbb{R})} \langle Z, R \rangle = \sup_{R \in \mathrm{SO}_{d}(\mathbb{R})} \langle D, R \rangle$$

We consider $R \in \mathrm{O}_{d}(\mathbb{R})$ with $\operatorname{det}(R) = -1$.

• [(a)] Show that there exists an orthonormal basis $(u_{1}, \ldots, u_{d})$ of $\mathbb{R}^{d}$ such that $Ru_{d} = -u_{d}$ and $u_{d}^{T} Rx = 0$ for all $x \in E_{1}$ where $E_{1} = \operatorname{Vect}(u_{1}, \ldots, u_{d-1})$.
• [(b)] Deduce that $R(E_{1}) \subset E_{1}$ and then that $R(E_{1}) = E_{1}$.

We consider $R \in \mathrm{O}_{d}(\mathbb{R})$ with $\operatorname{det}(R) = -1$, and an orthonormal basis $(u_{1}, \ldots, u_{d})$ of $\mathbb{R}^{d}$ such that $Ru_{d} = -u_{d}$ and $R(E_{1}) = E_{1}$ where $E_{1} = \operatorname{Vect}(u_{1}, \ldots, u_{d-1})$. We consider a matrix $D = \operatorname{Diag}(\alpha_{1}, \ldots, \alpha_{d}) \in \mathscr{M}_{d}(\mathbb{R})$ diagonal with diagonal entries $\alpha_{i} \geqslant 0$ in decreasing order. We denote $U = (u_{1} | \ldots | u_{d})$.

• [(a)] Verify that $\langle D, R \rangle = \langle S, R^{\prime} \rangle$ where $R^{\prime} = U^{T}RU$ and $S = U^{T}DU$.
• [(b)] Show that if $R_{0} = (R_{ij}^{\prime})_{1 \leqslant i,j \leqslant d-1} \in \mathscr{M}_{d-1}(\mathbb{R})$ then $R_{0} \in \mathrm{O}_{d-1}(\mathbb{R})$.

We consider $R \in \mathrm{O}_{d}(\mathbb{R})$ with $\operatorname{det}(R) = -1$, and the notation from question 23. We set $S_{0} = (S_{ij})_{1 \leqslant i,j \leqslant d-1} \in \mathscr{M}_{d-1}(\mathbb{R})$.

• [(a)] Show that $\langle D, R \rangle = \operatorname{tr}(S_{0} R_{0}) - S_{dd}$.
• [(b)] Show that $\operatorname{tr}(S_{0} R_{0}) \leqslant \operatorname{tr}(S_{0})$.
• [(c)] Show that $\operatorname{tr}(S_{0}) + S_{dd} = \operatorname{tr}(D)$ and deduce that $\langle D, R \rangle \leqslant \operatorname{tr}(D) - 2S_{dd}$.

For every matrix $M \in S_n(\mathbf{R})$ construct a vector subspace $F_M$ of $\mathcal{M}_{n,1}(\mathbf{R})$ of dimension $\pi(M)$ satisfying condition $(\mathcal{C}_M)$: $$\forall X \in F \setminus \{0_{n,1}\} \quad X^\top M X > 0.$$ We thus have $d(M) \geq \pi(M)$.